From $P(0,y)$ we deduce $f(0)=0$ or $f(y)=\frac{1}{2}$ for all $y$, which does not satisfy. So it must be the case that $f(0)=0$. Assuming $f$ is injective at $0$, we will prove $f(x)=x$ for all $x$. From $P(x,-x)$ we have $$-f\left(x^2\right)=\frac{xf(2x)}{2}+2xf\big(f(-x)\big).$$Adding $2f(x)f(x+y)$ to both sides and compare that to the equation gives $$f(x)f(x+y)=xf\big(f(y)\big)-xf\big(f(-x)\big)$$$$\Longrightarrow f(x)^2=-xf\big(f(-x)\big).$$From $P(x,0)$ we have $$2f(x)^2-f\left(x^2\right)=\frac{xf(2x)}{2}$$$$\Longrightarrow 2f(x)f(x+y)-f\left(x^2\right)=\frac{xf(2x)}{2}-2f(x)^2+2f(x)f(x+y).$$Comparing with the original equation again and then substitute $y\to0$ gives $f(x)^2=f\left(x^2\right)$ for all real $x$. So we have $$2f(x)^2-f\left(x^2\right)=f\left(x^2\right)=f\left((-x)^2\right)=\frac{xf(2x)}{2}=\frac{-xf(-2x)}{2}.$$Considering nonzero $x$ we have $f(2x)=-f(-2x)$, but since $f(0)=0$ then $f$ is odd. We also obtain $$f(x)^2=-xf\big(f(-x)\big)=\frac{xf(2x)}{2}\overset{f\text{ is odd}}{=}xf\big(f(x)\big)\Longrightarrow f(2x)=2f\big(f(x)\big).$$Now the given equation becomes $$2f(x)f(x+y)=2f(x)^2+xf(2y).$$Plugging $y\to-y$ and then adding both equations together gives $$2f(x)\big(f(x+y)+f(x-y)-2f(x)\big)=0.$$Since $f$ is injective at $0$, then from the last equality we deduce $f$ is additive. Since $f$ is non-negative for all non-negative $x$, we arrive at $f(x)=ax$, and we have $a=1$. So we may assume $f(b)=0$ for some nonzero $b$. Then by $P(b,0)$ we have $-f\left(b^2\right)=\frac{bf(2b)}{2}$, from this using $P(b,y)$ we have $f\big(f(y)\big)=0$ for all $y$. Now the equation becomes $$2f(x)f(x+y)=\frac{xf(2x)}{2}+f\left(x^2\right).$$If there exists a $c$ such that $f(c)\ne0$ then by $P(c,y)$ we have $f(y)=C$, but only $C=0$ is possible, which is a contradiction. So we are done.